A soccer ball is moving horizontally at a speed of 3.0 m/s. It then undergoes a constant negative acceleration. After 4.0 s, the ball is moving at 1.5 m/s. What is the ball’s displacement?
can you show me how you did it?
2006-09-25 16:31:48 · 4 answers · asked by tingerpoo 2 in Science & Mathematics ➔ Mathematics
One of the equations you should know for exams is that v^2=v0^2+2ad where v=velocity final, v0=initial velocity, d=distance travelled, and a=acceleration. You know v=1.5m/s, v0=3 m/s and you know that it decelerated by 1.5m/s in 4 seconds so a=-.375m/s^2. Plug and chug to find d=9m. Peace.
2006-09-25 16:40:12 · answer #1 · answered by g0atbeatr 3 · 1⤊ 0⤋
To find the acceleration we do:
a=deltav/t=-1.5/4=-.375
d=vot+1/2at^2
=3(4)+1/2(-.375)(4)^2=9m
2006-09-25 23:36:25 · answer #2 · answered by need help! 3 · 1⤊ 0⤋
Use the equation
(v2 - v1) = a t where a is the acceleration, t is time and v2 and v1 are the 2 velocities. This will give you the acceleration.
The displacement is (v2 + v1)/2 * t, basically the average velocity times the time.
2006-09-25 23:36:24 · answer #3 · answered by rscanner 6 · 0⤊ 1⤋
Vo = 3 m/s
V = 1.5 m/s
t = 4 s
a = ???
V = Vo + ½ at
1.5 = 3 + ½ * 4 * a
2a = - 1.5
a = -0.75 m/s²
d = Vo + ½ at²
d = 3 - ½ * ¾ * 16
d = 3 m
2006-09-26 00:37:57 · answer #4 · answered by M. Abuhelwa 5 · 0⤊ 0⤋